How Much Force Does a Gunshot Exert in Movie Scenes?

[zachwad47]

Homework Statement

Trying to find the total force felt by an average-weight man from his own gunshot plus a bullet incident on him, assuming negligible air resistance, static friction force, force lost to penetration, etc, based on the movie clip at . Have already calculated kinetic energy due to Harry's .44 round as appr 900 joules and that equal-and-opposite due to the baddie's shotgun blast as about 3300 joules, based on projectile weights of about .021 and .028 kg (.44 and shotgun) moving at around 300 and 490 m/s, respectively. This is giving me a total force felt (assuming negligible outside factors/energy loss and a totally elastic energy transfer) by the baddie of about 4300 joules... but from here, I am having a really hard time translating this to force or how much distance he *should* be thrown back. I have heard explanations about these sort of unrealistic gun knockbacks before and understand that the total force felt should never be enough to actually move an average-sized person any sizable distance (or the shooter EAO), but assuming a complete transfer of energy I am getting an output velocity which is even more unrealistic than the movie. I know I'm going wrong somewhere very simple but can't put my finger on it. Help!

Also guesstimating that he is knocked about 2m backwards over a 1s period...

Homework Equations

For Ek, f=ma, etc

The Attempt at a Solution

As above -- getting stuck!
Thanks for any advice you could give. This is what we midwesterners call a "brain fart"... I totally know how to do this but its just not coming to me. Do I have to apply a force only to a small bullet-sized area (again, assuming no penetration for simplicity's sake)?

 

[Delphi51]

Welcome to PF, Zach.
Just a couple of tips:
First, kinetic energy is not very helpful in finding the force. Certainly KE is not conserved in this situation.
The impulse formula F*Δt = m*Δv is probably your best bet. If you can find the change in velocity of the bullet in the gun and the time for the bullet to leave the gun, you will be able to find the force on bullet. The force on the bullet is the same as the force on the gun&man (Newton's 3rd law).

If you want to go straight to the backward speed of the gun and shooter, you might use conservation of momentum (initially zero).
p before = p after
0 = mv of bullet + mv of man&gun

 

[Andrew Mason]

As a matter of interest, bullet kinetic energy can be an important factor in the motion of the body if there is an explosive exit wound. This can actually cause the body to recoil (in the direction from which the bullet came - ie backward) with greater backward momentum than the forward momentum that the bullet was carrying. But in this case Delphi is correct: you have to apply conservation of momentum.

AM

 

[zachwad47]

Thanks all! I knew it was something simple :D